Dividing Pythagoras through by the hypotenuse squared

0:00Here are two equations. They look like the same kind of object and they are not. Two x plus three equals seven. That one is a question. It is asking which x, and there is exactly one answer. Run every value from minus five to five in halves through it - twenty one of them - and exactly one gets through. Now the second. x minus one, times x plus one, equals x squared minus one.

0:30Run the same twenty one values through that and all twenty one get through. It is not asking which x. It never constrains x at all. That is the difference, and it has a name. The second one is an identity. This video builds the first identity of trigonometry, and the point is that you do not have to memorise it. You already know the fact it comes from. Draw a right triangle. Put the right angle at B, with C on the left and A at the top.

1:04Mark the angle at A. That is the angle everything will be about. There is exactly one relation that every right triangle satisfies, whatever its size and whatever its shape. The square on AB plus the square on BC equals the square on AC. That is the whole input to this video. Nothing else goes in. And it really is a fact about the right angle rather than about the drawing.

1:34I built eight hundred and nineteen corners out of a few hundred triangles, some with a right angle in them and most without. Fifty two of those corners turned out to be right angles, measured rather than assumed. The relation held at those fifty two and at none of the other seven hundred and sixty seven. So it is not a fact about triangles. It is a fact about right angles.

2:02Now one move, and the move is the whole video. Divide every term by the square of AC. AC is the longest side, the one facing the right angle. The left side becomes AB squared over AC squared, plus BC squared over AC squared. The right side becomes AC squared over AC squared, which is one. Nothing has been added and nothing has been assumed. One equation has been divided by one quantity.

2:36But look at what the left side has turned into. AB over AC. A side that touches the angle at A, divided by the longest side. That is the cosine of A. BC over AC. The side facing the angle at A, divided by the longest side. That is the sine of A. Neither of those is new. Both were named the moment the six ratios were named. And a squared fraction is the fraction of the squares, so AB squared over AC squared is the cosine of A, squared.

3:14Same for the other one. That step is worth checking rather than accepting, so I checked it a different way. There is a way to compute the squared cosine at a corner using only the two arms leaving it, which never mentions a hypotenuse or a longest side at all. Across a hundred and four readings the two routes agreed at the corner they were both asked about. And at the other acute corner of the same triangle they disagreed - except in the eighteen readings where the two legs are equal, which is the one case where reading from the wrong corner cannot be caught.

3:57So put it together. The squared cosine of A plus the squared sine of A equals one. One line of notation before we go on, because the statement is unreadable without it. The exponent goes before the angle, not after it. That means the sine, then squared. It is not the sine of A squared, and it matters. Square each ratio and add, and you get one every time. Add first and then square, and on sixteen different triangles I got sixteen different answers, and not one of them was one.

4:36Look at the finished statement and notice what is missing. There is no length in it. No AB, no BC, no AC. A triangle went in and a triangle did not come out. That is the division doing its work, and it is worth watching directly. Take any of those triangles and scale it - double it, triple it, halve it. Over five hundred and twenty readings, every one of the three lengths changed and neither of the two ratios moved at all.

5:08Now stretch one leg instead, so the shape changes rather than the size. Over the same five hundred and twenty readings, the lengths changed and this time the ratios changed too. So the ratios are not indifferent to everything. They are indifferent to size. And here is the part that matters. The stretched triangles still have their right angle, so the two squares still add to one. The identity is holding for the angle's sake, not for the shape's.

5:42One question I have skipped over. Why divide by the square of AC? Why not divide by one of the legs? You can. It gives you a different statement - the next one in this module, in fact. I ran the same division routine over all one hundred and four right-angled readings with a leg underneath instead. The answer was never one. Not once. What it gives instead is the longest side over that leg, and the two squares over that leg add up to exactly that.

6:17So the choice of divisor is not cosmetic. Change it and you change the statement. The reason the longest side is the one to pick comes at the ends of the range, and that is where we are going next. The identity was true before any angle was chosen. But it is worth watching it land. At thirty degrees the sine is one half and the cosine is root three over two.

6:46Square them. A quarter, and three quarters. They add to one. At forty five degrees both ratios are one over root two. Square them. A half, and a half. They add to one. At sixty degrees the sine is root three over two and the cosine is one half. Square them. Three quarters and a quarter. One again. Three pairs that look nothing like each other, landing on the same answer.

7:14And the middle one is not taken on trust here either. A triangle with two equal legs produces a half and a half by measurement, eighteen times over. Now the two ends of the range, and be careful here. The claim is that the identity holds for every angle from zero through to a right angle, both ends included. But the derivation cannot reach either end, and I can show you that.

7:43Across every reading in the whole population, both squared ratios came out strictly between zero and one. Never zero, never one. That is because there is no right triangle with an angle of zero in it, and none with a second right angle. So at the ends there is no triangle to divide anything by, and the identity is not being derived. It is being checked. At zero degrees the squares are zero and one. They add to one.

8:17At a right angle they are one and zero. They add to one. The values at the ends were chosen, and this is a check that the choice was consistent. It also answers the question I left hanging. At each end, one leg has gone. Divide by the leg that has vanished and there is nothing to do - the division cannot be carried out at all. Divide by the longest side and it still works, at both ends.

8:46That is why the longest side is the safe thing to divide by. It is the one side that never disappears. Everything so far has used pretty numbers, and pretty numbers are misleading. So take a triangle with legs of twenty and twenty one. Nothing about those is special. But the third side comes out at twenty nine, which I checked rather than assumed. The sine is twenty over twenty nine and the cosine is twenty one over twenty nine.

9:20Square the first. Four hundred over eight hundred and forty one. Square the second. Four hundred and forty one over eight hundred and forty one. Four hundred plus four hundred and forty one is eight hundred and forty one. Which is exactly twenty nine squared. So the sum is eight hundred and forty one over eight hundred and forty one. One. Nothing memorable happened there, and that is the point. The identity is not living off the neatness of the special angles.

9:56Now what an identity is actually for. Rearrange it. The squared cosine is one minus the squared sine. So if you know the sine, you know the cosine - up to a sign. Up to a sign, because two numbers have the same square. On sixteen triangles I took one minus the squared sine and asked for its two square roots. Every time there were two, and they added to zero.

10:24Every time, the one the drawing actually shows was the positive one. Which makes sense. The cosine here is one length divided by another, and lengths are not negative. So the negative root is discarded, for the same reason a negative length was discarded earlier. Now watch what that buys. Suppose you are told only that the sine is one half. Then the squared sine is a quarter, so the squared cosine is three quarters.

10:57The squared tangent is a quarter over three quarters, which is a third. And the squared secant is one over three quarters, which is four thirds. Three quarters, a third, four thirds. That is the thirty degree column, and it came out of one number and one identity. One last thing, because this is where identities earn their keep. Take one plus the sine, times one minus the sine, times the squared secant.

11:28Three brackets, nothing in common, no reason to expect anything tidy. The first two are a sum and a difference, so they multiply to one minus the squared sine. And now the identity, used once. One minus the squared sine IS the squared cosine. The squared secant is one over the squared cosine. So the whole thing is the squared cosine over the squared cosine. One. I ran that product on all sixteen of the ugly triangles and it came to one every single time.

12:07Put the wrong reciprocal underneath instead and it collapsed on none of them. So the collapse is the identity's doing and not the arithmetic's good luck. And notice where the identity entered. Exactly one step, right in the middle. That is the shape of almost every proof in this part of the subject. Convert to sines and cosines, find the one place where a difference of one and a square is sitting, and replace it.

12:38All of which came from one right triangle and one division.